= Solution
For this curve, the square-class image in the first <two-isogeny descent> is contained in $\{1,-1,11,-11\}$. All four classes occur: $Q=(-1,1)$ gives $-1$, $T=(0,0)$ gives $11$, and $Q+T$ gives $-11$. The isogenous curve is
$$
E':y^2=x^3-26x^2+125x.
$$
Its image is contained in $\{1,-1,5,-5\}$. Negative $x$ cannot occur because $x(x^2-26x+125)<0$ for $x<0$, while $1$ and $5$ are represented by the identity and $(0,0)$. Thus the two image orders are four and two, and
$$
2^{r+2}=4\cdot2
$$
gives $r=1$.
The point $P$ has order three because $2P=-P$, and $T$ has order two, so the rational torsion contains a cyclic subgroup of order six. At the good primes $7$ and $13$, direct point counting gives
$$
\#E(\mathbb F_7)=12,
\qquad \#E(\mathbb F_{13})=18.
$$
Reduction bounds the rational torsion order by their greatest common divisor, namely six, so this is all the torsion. The <structure theorem for finitely generated modules over a principal ideal domain> now gives
$$
E(\mathbb Q)\cong\mathbb Z/6\mathbb Z\times\mathbb Z.
$$
Thus one may take $d_1=1$, $d_2=6$, and $r=1$.
Solved by gpt-5.6-sol high.
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